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Apumpis used toraisethe pressure of a fluid, and it operatesby converting the mechanical shaft work to mechanicalenergyof the fluid flow. A gasoline pump is measured to consume 3.5kW ofmechanical shaftpower when operating. The diameters of the outlet and inlet of the pump are the same and the pressuredifferential between them is 7 kPa. Determine the maximum possible volumeflow rateof gas

Respuesta :

Answer:

maximum possible volume flow rate = V = 0.5m^3/s

Explanation:

Given power consumed = 3.5kW

pressure difference ( delta -P) = 7kPa

let maximum possible volume flow rate = V

The rate of flow of work = W = V x Delta -P

hence, V = W/delta P

V = 3.5/7

V = 0.5m^3/s